# How do you implicitly differentiate (cos(x^2+y^2))=(e^xy) ?

$y ' = \frac{- y \cdot {e}^{x} - 2 x \cdot \sin \left({x}^{2} + {y}^{2}\right)}{{e}^{x} + 2 y \cdot \sin \left({x}^{2} + {y}^{2}\right)}$

#### Explanation:

the given equation is
$\cos \left({x}^{2} + {y}^{2}\right) = {e}^{x} \cdot y$

$\frac{d}{\mathrm{dx}} \cos \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({e}^{x} \cdot y\right)$

$- \sin \left({x}^{2} + {y}^{2}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(y\right) + y \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$

$- \sin \left({x}^{2} + {y}^{2}\right) \cdot \left(2 {x}^{2 - 1} + 2 {y}^{2 - 1} \cdot y '\right) = {e}^{x} \left(y '\right) + y \left({e}^{x}\right)$

$- y \cdot {e}^{x} - 2 x \cdot \sin \left({x}^{2} + {y}^{2}\right) = {e}^{x} \cdot y ' + 2 y \cdot \sin \left({x}^{2} + {y}^{2}\right) \cdot y '$

$- y \cdot {e}^{x} - 2 x \cdot \sin \left({x}^{2} + {y}^{2}\right) = \left({e}^{x} + 2 y \cdot \sin \left({x}^{2} + {y}^{2}\right)\right) \cdot y '$

(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))=((e^x+2y*sin(x^2+y^2))*y')/((e^x+2y*sin(x^2+y^2))

(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))=cancel((e^x+2y*sin(x^2+y^2))*y')/(cancel(e^x+2y*sin(x^2+y^2))

$y ' = \frac{- y \cdot {e}^{x} - 2 x \cdot \sin \left({x}^{2} + {y}^{2}\right)}{{e}^{x} + 2 y \cdot \sin \left({x}^{2} + {y}^{2}\right)}$

God bless....I hope the explanation is useful.